0
$\begingroup$

Let $E/K$ be a field extension, $a,b\in E$ algebraic over $K$. Show: $\text{Min}(a,K,X)$ irreducible over $K(b)$ if and only if $\text{Min}(b,K,X)$ irreducible over $K(a)$

My attemp:

(1) Let deg $\text{Min}(a,K,X)$ = n, deg $\text{Min}(b,K,X)$ = m. We know: $[F(ab):F] = [F(ab):F(a)]\cdot [F(a):F] = [F(ab):F(a)]\cdot n $ $[F(ab):F] = [F(ab):F(b)]\cdot [F(b):F] = [F(ab):F(b)]\cdot m $

$\Rightarrow [F(ab):F(a)]\cdot n = [F(ab):F(b)]\cdot m$

Let $f:= \text{Min}(a,K,X)$ irreducible over $K(b)$. Since f is minimal Polynomial of $a$ over $K[X]$, it is the unique normalized irreducible polynomial with $f(a)=0$. Let $f'(a)=0$ and $f'(X)\in F(b)[X]$ be the minimal polynomial of $a$ over $K(b)$. Since $f\in K[X]$, also $f\in K(b)[X]$ and $\text{deg}\, f'\leq \text{deg}\, f$. Because $f:= \text{Min}(a,K,X)$ irreducible over $K(b) \Rightarrow f'=f \Rightarrow \text{deg}\, f'= \text{deg}\, f\Rightarrow [F(ab):F(b)]=n $.

Is this ok?

$\endgroup$
1
$\begingroup$

Your proof appears OK, except that

  • you switch from $K$ to $F$ and back (that's of course no serious problem)
  • you want to write $F(a, b)$ (or $K(a, b) $) instead of $F(ab) $

Also note that if you use the arguments of this answer, you'll see that your conditions are both equivalent to the symmetric condition $K(a) \cap K(b) = K$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.